Let \(A\) be linear on a vector space \(W\) over a field \(k\), with \(A^5=I\).
If \(\operatorname{char}k\ne5\), show that \(W=U\oplus V\), where \(U=\{u:Au=u\}\) and \(AV=V\).
Give an example showing this can fail in characteristic \(5\).
Solution
1. Assume\[ \operatorname{char}k\ne5 \]and define\[ P \coloneqq \frac15(I+A+A^2+A^3+A^4). \]Then\[ \operatorname{im}P=U. \]
Since \(5\) is nonzero in \(k\), it is invertible and \(P\) is well-defined. For every \(w\in W\), \[ \begin{aligned} APw &= \frac15(Aw+A^2w+A^3w+A^4w+A^5w)\\ &= \frac15(Aw+A^2w+A^3w+A^4w+w)\\ &= Pw. \end{aligned} \] Thus \(Pw\in U\), so \[ \operatorname{im}P\subseteq U. \]
Conversely, if \(u\in U\), then \(Au=u\), hence \[ Pu = \frac15(u+u+u+u+u) = u. \] Therefore every \(u\in U\) belongs to \(\operatorname{im}P\), proving equality.
2. The operator \(P\) is a projection:\[ P^2=P. \]
By step 1, \(Pw\in U\) for every \(w\in W\). The same step shows that \(P\) acts as the identity on \(U\). Hence \[ P^2w=P(Pw)=Pw \] for every \(w\).
3. If\[ V\coloneqq\ker P, \]then\[ W=U\oplus V. \]
For every \(w\in W\), \[ w = Pw+(w-Pw). \] By step 1, \[ Pw\in U, \] and by step 2, \[ P(w-Pw) = Pw-P^2w = 0, \] so \(w-Pw\in V\). Thus \(W=U+V\).
If \(x\in U\cap V\), then step 1 gives \[ Px=x, \] while \(x\in V\) gives \(Px=0\). Hence \(x=0\), so the sum is direct.
4. The complement \(V\) satisfies\[ AV=V. \]
The operator \(P\) is a polynomial in \(A\), so \[ PA=AP. \] If \(v\in V\), then \[ P(Av) = A(Pv) = 0, \] so \(Av\in V\). Hence \[ AV\subseteq V. \]
Also \[ A^{-1}=A^4 \] because \(A^5=I\). Since \(P\) also commutes with \(A^4\), the same argument gives \[ A^{-1}V\subseteq V. \] Applying \(A\) yields \[ V\subseteq AV. \] Therefore \(AV=V\). This proves part (a).
5. For part (b), let\[ k=\FF_5, \qquad W=k^2, \]with basis \(e_1,e_2\), and define\[ Ae_1=e_1, \qquad Ae_2=e_1+e_2. \]Then\[ A^5=I. \]
Write \[ A=I+N, \] where \[ Ne_1=0, \qquad Ne_2=e_1. \] Then \(N^2=0\). Hence in characteristic \(5\), \[ A^5 =(I+N)^5 = I+5N = I. \]
6. For the operator in step 5,\[ U=\operatorname{span}(e_1). \]
For \[ v=ae_1+be_2, \] one has \[ Av=(a+b)e_1+be_2. \] Thus \(Av=v\) exactly when \(b=0\), which gives the stated fixed space.
7. The fixed space \(U\) in step 6 has no \(A\)-invariantcomplement.
Any complement \(V\) of the one-dimensional subspace \(U\) in \(W=k^2\) must itself be one-dimensional. Write \[ V=\operatorname{span}(ae_1+be_2) \] with \(b\ne0\), since otherwise \(V=U\).
If \(AV=V\), then for some \(\lambda\in k\), \[ A(ae_1+be_2) = \lambda(ae_1+be_2). \] Using step 6, this becomes \[ (a+b)e_1+be_2 = \lambda ae_1+\lambda be_2. \] Comparing the \(e_2\) coefficients and using \(b\ne0\) gives \[ \lambda=1. \] Comparing the \(e_1\) coefficients then gives \[ a+b=a, \] so \(b=0\), a contradiction. Therefore no such invariant complement exists.
8. Q.E.D.
Steps <1>1–<1>4 prove part (a), and steps <1>5–<1>7 give the required counterexample for part (b).